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Richard Cleve DC 3524 cleve@cs.uwaterloo.ca
Introduction to Quantum Information Processing CS 467 / CS 667 Phys 667 / Phys 767 C&O 481 / C&O 681 Lecture 2 (2005) Richard Cleve DC 3524
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Superdense coding
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How much classical information in n qubits?
2n1 complex numbers apparently needed to describe an arbitrary n-qubit pure quantum state: 000000 + 001001 + 010010 + + 111111 Does this mean that an exponential amount of classical information is somehow stored in n qubits? Not in an operational sense ... For example, Holevo’s Theorem (from 1973) implies: one cannot convey more than n classical bits of information in n qubits
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Holevo’s Theorem U U ψ ψ
Easy case: Hard case (the general case): U ψ n qubits b1 b2 b3 bn ψ n qubits b1 b2 b3 bn U 0 m qubits bn+1 bn+2 bn+3 bn+4 bn+m b1b bn certainly cannot convey more than n bits! The difficult proof is beyond the scope of this course
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Superdense coding (prelude)
Suppose that Alice wants to convey two classical bits to Bob sending just one qubit ab Alice Bob ab By Holevo’s Theorem, this is impossible
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Superdense coding ab ab Alice Bob
In superdense coding, Bob is allowed to send a qubit to Alice first ab Alice Bob ab How can this help?
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How superdense coding works
Bob creates the state 00 + 11 and sends the first qubit to Alice Alice: if a = 1 then apply X to qubit if b = 1 then apply Z to qubit send the qubit back to Bob ab state 00 00 + 11 01 00 − 11 10 01 + 10 11 01 − 10 Bell basis Bob measures the two qubits in the Bell basis
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Measurement in the Bell basis
Specifically, Bob applies input output 00 + 11 00 01 + 10 01 00 − 11 10 01 − 10 11 H to his two qubits ... and then measures them, yielding ab This concludes superdense coding
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Teleportation
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Incomplete measurements (I)
Measurements up until now are with respect to orthogonal one-dimensional subspaces: The orthogonal subspaces can have other dimensions: 0 1 2 span of 0 and 1 2 (qutrit)
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Incomplete measurements (II)
Such a measurement on 0 0 + 1 1 + 2 2 (renormalized) results in 00 + 11 with prob 02 + 12 2 with prob 22
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Measuring the first qubit of a two-qubit system
0000 + 0101 + 1010 + 1111 Defined as the incomplete measurement with respect to the two dimensional subspaces: span of 00 & 01 (all states with first qubit 0), and span of 10 & 11 (all states with first qubit 1) Result is the mixture 0000 + 0101 with prob 002 + 012 1010 + 1111 with prob 102 + 112
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Easy exercise: show that measuring the first qubit and then measuring the second qubit gives the same result as measuring both qubits at once
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Teleportation (prelude)
Suppose Alice wishes to convey a qubit to Bob by sending just classical bits 0 + 1 0 + 1 If Alice knows and , she can send approximations of them ―but this still requires infinitely many bits for perfect precision Moreover, if Alice does not know or , she can at best acquire one bit about them by a measurement
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Teleportation scenario
In teleportation, Alice and Bob also start with a Bell state (1/2)(00 + 11) 0 + 1 and Alice can send two classical bits to Bob Note that the initial state of the three qubit system is: (1/2)(0 + 1)(00 + 11) = (1/2)(000 + 011 + 100 + 111)
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How teleportation works
(0 + 1)(00 + 11) (omitting the 1/2 factor) = 000 + 011 + 100 + 111 = ½(00 + 11)(0 + 1) + ½(01 + 10)(1 + 0) + ½(00 − 11)(0 − 1) + ½(01 − 10)(1 − 0) Initial state: Protocol: Alice measures her two qubits in the Bell basis and sends the result to Bob (who then “corrects” his state)
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What Alice does specifically
Alice applies to her two qubits, yielding: ½00(0 + 1) + ½01(1 + 0) + ½10(0 − 1) + ½11(1 − 0) (00, 0 + 1) with prob. ¼ (01, 1 + 0) with prob. ¼ (10, 0 − 1) with prob. ¼ (11, 1 − 0) with prob. ¼ Then Alice sends her two classical bits to Bob, who then adjusts his qubit to be 0 + 1 whatever case occurs
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Bob’s adjustment procedure
Bob receives two classical bits a, b from Alice, and: if b = 1 he applies X to qubit if a = 1 he applies Z to qubit 00, 0 + 1 01, X(1 + 0) = 0 + 1 10, Z(0 − 1) = 0 + 1 11, ZX(1 − 0) = 0 + 1 yielding: Note that Bob acquires the correct state in each case
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Summary of teleportation
0 + 1 H Alice a 00 + 11 b Bob 0 + 1 X Z Quantum circuit exercise: try to work through the details of the analysis of this teleportation protocol
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No-cloning theorem
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Classical information can be copied
a a What about quantum information? ? ψ 0
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Candidate: works fine for ψ = 0 and ψ = 1 ... but it fails for ψ = (1/2)(0 + 1) ... ... where it yields output (1/2)(00 + 11) instead of ψψ = (1/4)(00 + 01 + 10 + 11)
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No-cloning theorem U ψ 0 g ψψ′ = ψψ′ψψ′gg′ so
Theorem: there is no valid quantum operation that maps an arbitrary state ψ to ψψ Proof: Let ψ and ψ′ be two input states, yielding outputs ψψg and ψ′ψ′g′ respectively ψ 0 g U Since U preserves inner products: ψψ′ = ψψ′ψψ′gg′ so ψψ′(1− ψψ′gg′) = 0 so ψψ′ = 0 or 1
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THE END
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